24. Balkanska matematiqka olimpijada

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "24. Balkanska matematiqka olimpijada"

Transcription

1 4. Balkanska matematika olimpijada Rodos, Gka 8. apil U konveksnom etvoouglu ABCD vaжi AB = BC = CD, dijagonale AC i BD su azliite duжine i seku se u taki E. Dokazati da je AE = DE ako i samo ako je BAD+ ADC = 10. (Albanija). Na i sve funkcije f : R R takve da za sve ealne bojeve x, y vaжi f(f(x) + y) = f(f(x) y) + 4f(x)y. (Bugaska) 3. Na i sve piodne bojeve n za koje postoji pemutacija σ bojeva 1,,..., n takva da je boj σ(1) + σ() + + σ(n) acionalan. (Sbija) 4. Dat je ceo boj n 3. Neka su C 1, C i C 3 ganice ti konveksna n-tougla u avni takva da je pesek svake dve od njih konaan skup taaka. Na i najve i mogu i boj taaka skupa C 1 C C 3. (Tuska)

2 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas, 8 Apil 007 Poblem 1. Let ABCD be a convex uadilateal with AB = BC = CD, AC BD and let E be the intesection point of its diagonals. Pove that AE = DE if and only if BAD + ADC = 10. Poblem. Find all functions f : such that ( ( ) ) ( ( ) ) ( ) f f x + y = f f x y + 4 f x y, fo any xy,. Poblem 3. Find all positive integes n such that thee is a pemutation σ of the set {1,,,n} fo which σ(1) + σ() + + σ( n) is a ational numbe. Note: A pemutation of the set {1,,,n} is a one-to-one function of this set to itself. Poblem 4. Fo a given positive intege n >, let C1, C, C 3 be the boundaies of thee convex n-gons in the plane such that C C, C C, C C ae finite. Find the maximum numbe of points of the set C C C 3. 1 Time allowed 4 hous and 30 minutes Each poblem is woth 10 points.

3

4 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem. Find all functions f : R! R such that f(f(x) + y) = f(f(x) y) + 4f(x)y fo any x; y R: Solution. It is clea that the function f 0 satis es the given condition. y Assume that f 6 0: Choose x 0 such that f(x 0 ) 6= 0 and set ey = 4f(x 0 ) y: Then plugging in to the given elation x 0 fo x and ey fo y we get fo any y = f(f(x 0 ) + ey) f(f(x 0 ) ey): (1) Fo any y 1 ; y, plugging in to (1) y 1 y fo y, we get y 1 y y = f(f(x 0 ) + ^ 1 y ) f(f(x 0 ) y^ 1 y ): In othe wods, fo any y 1 ; y thee exist x 1 (y 1 ; y ) = f(x 0 ) + ]y 1 y ; x (y 1 ; y ) = f(x 0 ) + ]y 1 y such that y 1 y = f(x 1 ) f(x ), i.e. such that f(x 1 ) y 1 = f(x ) y : () On the othe hand, eplacing y by f(x) y in the given condition gives f(f(x) y) = f(y) + 4f(x)(f(x) y); i.e. f(y) y = f(f(x) y) ((f(x) y) : (3) Now if fo any two y 1 ; y, we plug in to (3), x 1 (y 1 ; y ) and x (y 1 ; y ) espectively, we get f(y 1 ) y 1 = f(f(x 1 ) y 1 ) ((f(x 1 ) y 1 ) and f(y ) y = f(f(x ) y ) ((f(x ) y ) Then by () we get f(y 1 ) y1 = f(y ) y: Since this happens fo any two y 1 ; y, we conclude that f(x) x = constant fo all x, thus f(x) = x + c; c R. It is easy to check that such a function satis es the condition of the poblem. 1

5 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 3. Find all positive s integes n such that thee is a pemutation of the set f1; ; :::; ng; fo which (1) + () + ::: + p (n) is a ational numbe. Note: A pemutation of the set f1; ; :::; ng is a one-to-one function of this set to itself. s Solution. Fo some n N, let (1) + () + ::: + p (n) = 1 Q. Suaing both sides of the euation we get that () + (3) + ::: + p (n) is s also ational. s Using the same easoning ecusively, we get that fo evey k f1; : : : ; ng, (k) + (k + 1) + ::: + p (n) is ational as well. Knowing that the suae oot of a positive intege is eithe intege o iational, we have that p s (n) is intege. Similaly, we get that (k) + (k + 1) + ::: + p (n), fo evey k f1; : : : ; ng, is intege. Note that s fo k = 1 we get 1 N. We de ne a k as a k = n + n + ::: + p n, fo all k 1. It is easy to {z } k pove by induction that a k < p s n + 1, fo evey k 1. Theefoe, we have (1) + () + ::: + p (n) < a n < p n + 1, implying 1 < p n + 1. Let ` be the positive intege that satis es ` n < (`+1). Fo some i; 1 i n, we have (i) = `. We distinguish two cases: Fist case: i 6= n: Then we have ` < s ` + (i + 1) + ::: + p (n) < p n+1 < `+, implying 1

6 s ` + (i + 1) + ` + 1 = ::: + p (n) = ` + 1. But then it follows that giving ` < 1. A contadiction. Second case: i = n: (i + 1) + ::: + p (n) < p n + 1 < ` + ; Fo ` > 1, ` 1 belongs to the set f(1); :::; (n 1)g. Let j < n be such that (j) = ` 1. Similaly to the st case, we have implying ` < s ` 1 + ` 1 + ` + = (j + 1) + ::: + p` < p n + 1 < ` + ; p (j + 1) + ::: + p` = ` + 1, and (j + 1) + ::: + p` < p n + 1 < ` + ; a contadiction. If ` = 1, then n f1; ; 3g. Checking though all the possibilities, it is easy to see that fo n = 1 and n = 3 thee exist pemutations that satisfy the initial condition. Namely, fo n = 1 we have p 1 = 1, and fo n = 3, we have + p 3 + p 1 =. Fo n = thee is no such pemutation.

7 4 th BALKAN MATHEMATICAL OLYMPIAD Rhodes, Hellas (Apil 8, 007) Poblem 4. Fo a given positive intege n >, let C 1 ; C ; C 3 be the boundaies of thee convex n-gons in the plane such that the sets C 1 \ C, C \ C 3, C 3 \ C 1 ae nite. Find the maximum numbe of points of the set C 1 \ C \ C 3. Solution 1: Let us st obseve that, if a line intesects a convex n gon at nitely many points, then the numbe of such points is at most. Theefoe any two of the n gons may intesect in at most n points. Choose two of the n gons, C 1 ; C, and say that thei intesection points ae p 1 ; p ; : : : ; p k. Thus k n. Say that the union of the set of vetices of C 1 and C is f 1 ; ; : : : ; n g. We note that it is possible to have i = j fo some i 6= j. We will de ne a one-to-one function f fom fp 1 ; p ; : : : ; p k g to f 1 ; ; : : : ; n g as follows. Fist of all, oient all n gons in the clockwise diection. Thus, if one taveses an n gon accoding to this oientation, the inteio is on the ight and the exteio is on the left. Fo evey p i, thee exist pecisely two line segments (of non-zeo length) which ae subsets of C 1 o C, say [ j ; p i ] on C 1 and [ k ; p i ] on C, such that one can appoach to p i via these line segments in the clockwise diection. Suppose, that the two vectos p i j and p i k, in this ode, fom a ight handed coodinate system. Then none of the points on [ j ; p i ] can be on o in the inteio of C, since fo any point on o in the inteio of C, the vectos p i and p i k ae eithe positive multiples of each othe, o fom a left handed coodinate system. In this case we set f(p i ) = j. Othewise we set f(p i ) = k. In both cases, the agument above shows that thee ae no othe intesection points between f(p i ) and p i, in the clockwise diection. Let us now show that f is 1 1. If f(p i ) = f(p l ) = and say (without loss of geneality) belong to C 1, then the st intesection point encounteed when one stats fom and taveses C 1 in the clockwise diection has to be both p i and p l, hence p i = p l. Now let us estimate the numbe of p i s that can be contained by the thid polygon C 3. Each edge of C 3 contains exactly 0; 1 o of the p i s. Suppose that a given edge of C 3 contains of the p i s, say p 1 and p. Since C 1 and C ae convex and thei intesection with C 3 is geneic, they should have vetices between (in the clockwise sense) p 1 and p (with this ode), and outside C 3. We claim that at least one of these vetices is not in the set f(c 1 \ C \ C 3 ). Let 1 C 1 and C espectively be between (in the clockwise sense) p 1 and p (with this ode). If 1 is on o in the inteio of C (o is on o in the inteio of C 1 ), then 1 (o ) is not in the image of f, since ecall that f(p) fo any p C 1 \ C (thus fo any p C 1 \ C \ C 3 ) is a point of one of C 1 ; C not on o in the inteio of the othe. So the claim is established in this case. The emaining case is to assume that none of the vetices of any of C 1 ; C that lie outside C 3 (and between (in the clockwise sense) p 1 and p 1

8 (with this ode)) also lies in the inteio of the othe of C 1 ; C. In this case clealy the polygons C 1 and C must meet at some point between (in the clockwise sense) p 1 and p (with this ode). Say p 3 is the closest to p 1 such point. Then clealy f(p 3 ) is a vetex of one of C 1 ; C between (in the clockwise sense) p 1 and p (with this ode). These pats of C 1 ; C though, lie outside C 3 ; the inteio of C 3 lie on the othe side of the line p 1 p. Thus f(p 3 ) is not in f(c 1 \ C \ C 3 ) and the claim is established in all cases. Fo the side a of C 3 containing p 1 ; p let us call (a) a vetex as the one in the claim we just poved. It is easy to see that fo distinct sides a; b of C 3 that contain two of the p s, the points a ; b ae distinct. Indeed, let a contain p 1 ; p and b contain p 3 ; p 4 among the p s. If one of a ; b belongs to one of C 1 ; C and the othe does not belong to it, we ae okay. If both a ; b belong to say C 1, then in a clockwise tou aound C 1 stating at p 1, we meet p 1 ; p ; p 3 ; p 4 in this ode. If not, say the ode is p 1 ; p 3 ; p ; p 4. Then the segments p 1 p, p 3 ; p 4 intesect at an inteio point, since C 1 is a convex polygon. But then the sides a; b of C 3 have a common inteio point, a contadiction. So the coect ode is p 1 ; p ; p 3 ; p 4. But we know that adding (a); (b) in this tou the coect ode is p 1 ; (a); p ; p 3 ; (b); p 4. Thus (a); (b) ae distinct as claimed. Now if x of the edges of C 3 contain 1 of the p i s and y of them contain of the p i s, then x + y numbe of sides of C 3, i.e. x + y n. The numbe of points in C 1 \ C \ C 3 is x + y. Since f is injective, x + y is also the numbe of s in f(c 1 \ C \ C 3 ). Also, by the agument in the pevious paagaph, we see that fo evey distinct edge of C 3 containing points we can assign a coesponding distinct i outside the image of f(c 1 \ C \ C 3 ). Theefoe x is less o eual to the numbe of s that do not belong in f(c 1 \ C \ C 3 ). So (x + y) + y is at most as much as the numbe of s. I.e. x + 3y n. Adding this with x + y n and dividing by, and also taking into account that x + y is an intege x + y b 3n c Let us now show that this is the best uppe bound fo evey n 3. One way (among many) to constuct an example is as follows: Constuct two egula n gons C 1 ; C with the same cente, such that thei intesection points fom a egula n gon. Call the vetices p 1 ; p ; : : : ; p n in a cyclic ode. Let the cicumcicle of this n gon be C. Then let the n gon bounded by the lines p 1 p 3 ; p 5 p 7 ; p 9 p 11 ; : : : (including p k+1 p 1 in case n is an odd n = k + 1) togethe with the tangent lines to C at p 4 ; p 8 ; p 1 ; : : : be C 3. It can easily be checked that jc 1 \ C \ C 3 j = b 3nc. Solution : Let A and B be two conseutive points of C 1 \ C \ C 3 obseved in the clockwise diection fom a point in the inteio of all thee n-gons. Let s look fo each C i its section in the clockwise diection between A and B exluding these points. If some two of these sections both do not contain any vetices of thei coesponding n-gons, then the segment AB belongs to both n-gons, a contadiction.

9 Thus at least two of these segments have at least one vetex each, and moeovethey do not contain the segment. Tivially, two distinct such vetices exist. Since thee exist jc 1 \ C \ C 3 j many conseutive points points A and B of C 1 \ C \ C 3, thee should exist at least jc 1 \ C \ C 3 j distinct vetices of the thee n-gons. Thus jc 1 \ C \ C 3 j 3n i.e. jc 1 \ C \ C 3 j b 3nc since (jc 1 \ C \ C 3 j is an intege as well). Actually we can achieve this uppe bound by the example given in the Solution 1. 3

of the contestants play as Falco, and 1 6

of the contestants play as Falco, and 1 6 JHMT 05 Algeba Test Solutions 4 Febuay 05. In a Supe Smash Bothes tounament, of the contestants play as Fox, 3 of the contestants play as Falco, and 6 of the contestants play as Peach. Given that thee

More information

COLLAPSING WALLS THEOREM

COLLAPSING WALLS THEOREM COLLAPSING WALLS THEOREM IGOR PAK AND ROM PINCHASI Abstact. Let P R 3 be a pyamid with the base a convex polygon Q. We show that when othe faces ae collapsed (otated aound the edges onto the plane spanned

More information

A proof of the binomial theorem

A proof of the binomial theorem A poof of the binomial theoem If n is a natual numbe, let n! denote the poduct of the numbes,2,3,,n. So! =, 2! = 2 = 2, 3! = 2 3 = 6, 4! = 2 3 4 = 24 and so on. We also let 0! =. If n is a non-negative

More information

Stanford University CS259Q: Quantum Computing Handout 8 Luca Trevisan October 18, 2012

Stanford University CS259Q: Quantum Computing Handout 8 Luca Trevisan October 18, 2012 Stanfod Univesity CS59Q: Quantum Computing Handout 8 Luca Tevisan Octobe 8, 0 Lectue 8 In which we use the quantum Fouie tansfom to solve the peiod-finding poblem. The Peiod Finding Poblem Let f : {0,...,

More information

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) , R Pena Towe, Road No, Contactos Aea, Bistupu, Jamshedpu 8, Tel (657)89, www.penaclasses.com IIT JEE Mathematics Pape II PART III MATHEMATICS SECTION I Single Coect Answe Type This section contains 8 multiple

More information

Chapter 3: Theory of Modular Arithmetic 38

Chapter 3: Theory of Modular Arithmetic 38 Chapte 3: Theoy of Modula Aithmetic 38 Section D Chinese Remainde Theoem By the end of this section you will be able to pove the Chinese Remainde Theoem apply this theoem to solve simultaneous linea conguences

More information

FREE Download Study Package from website: &

FREE Download Study Package from website:  & .. Linea Combinations: (a) (b) (c) (d) Given a finite set of vectos a b c,,,... then the vecto xa + yb + zc +... is called a linea combination of a, b, c,... fo any x, y, z... R. We have the following

More information

15 Solving the Laplace equation by Fourier method

15 Solving the Laplace equation by Fourier method 5 Solving the Laplace equation by Fouie method I aleady intoduced two o thee dimensional heat equation, when I deived it, ecall that it taes the fom u t = α 2 u + F, (5.) whee u: [0, ) D R, D R is the

More information

EM Boundary Value Problems

EM Boundary Value Problems EM Bounday Value Poblems 10/ 9 11/ By Ilekta chistidi & Lee, Seung-Hyun A. Geneal Desciption : Maxwell Equations & Loentz Foce We want to find the equations of motion of chaged paticles. The way to do

More information

PROBLEM SET #1 SOLUTIONS by Robert A. DiStasio Jr.

PROBLEM SET #1 SOLUTIONS by Robert A. DiStasio Jr. POBLM S # SOLUIONS by obet A. DiStasio J. Q. he Bon-Oppenheime appoximation is the standad way of appoximating the gound state of a molecula system. Wite down the conditions that detemine the tonic and

More information

S7: Classical mechanics problem set 2

S7: Classical mechanics problem set 2 J. Magoian MT 9, boowing fom J. J. Binney s 6 couse S7: Classical mechanics poblem set. Show that if the Hamiltonian is indepdent of a genealized co-odinate q, then the conjugate momentum p is a constant

More information

So, if we are finding the amount of work done over a non-conservative vector field F r, we do that long ur r b ur =

So, if we are finding the amount of work done over a non-conservative vector field F r, we do that long ur r b ur = 3.4 Geen s Theoem Geoge Geen: self-taught English scientist, 793-84 So, if we ae finding the amount of wok done ove a non-consevative vecto field F, we do that long u b u 3. method Wok = F d F( () t )

More information

VECTOR MECHANICS FOR ENGINEERS: STATICS

VECTOR MECHANICS FOR ENGINEERS: STATICS 4 Equilibium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Fedinand P. Bee E. Russell Johnston, J. of Rigid Bodies Lectue Notes: J. Walt Ole Texas Tech Univesity Contents Intoduction Fee-Body Diagam

More information

Analysis of simple branching trees with TI-92

Analysis of simple branching trees with TI-92 Analysis of simple banching tees with TI-9 Dušan Pagon, Univesity of Maibo, Slovenia Abstact. In the complex plane we stat at the cente of the coodinate system with a vetical segment of the length one

More information

The geometric construction of Ewald sphere and Bragg condition:

The geometric construction of Ewald sphere and Bragg condition: The geometic constuction of Ewald sphee and Bagg condition: The constuction of Ewald sphee must be done such that the Bagg condition is satisfied. This can be done as follows: i) Daw a wave vecto k in

More information

Permutations and Combinations

Permutations and Combinations Pemutations and Combinations Mach 11, 2005 1 Two Counting Pinciples Addition Pinciple Let S 1, S 2,, S m be subsets of a finite set S If S S 1 S 2 S m, then S S 1 + S 2 + + S m Multiplication Pinciple

More information

Geometry Unit 4b - Notes Triangle Relationships

Geometry Unit 4b - Notes Triangle Relationships Geomety Unit 4b - Notes Tiangle Relationships This unit is boken into two pats, 4a & 4b. test should be given following each pat. quidistant fom two points the same distance fom one point as fom anothe.

More information

Deterministic vs Non-deterministic Graph Property Testing

Deterministic vs Non-deterministic Graph Property Testing Deteministic vs Non-deteministic Gaph Popety Testing Lio Gishboline Asaf Shapia Abstact A gaph popety P is said to be testable if one can check whethe a gaph is close o fa fom satisfying P using few andom

More information

Green s Identities and Green s Functions

Green s Identities and Green s Functions LECTURE 7 Geen s Identities and Geen s Functions Let us ecall The ivegence Theoem in n-dimensions Theoem 7 Let F : R n R n be a vecto field ove R n that is of class C on some closed, connected, simply

More information

THE JEU DE TAQUIN ON THE SHIFTED RIM HOOK TABLEAUX. Jaejin Lee

THE JEU DE TAQUIN ON THE SHIFTED RIM HOOK TABLEAUX. Jaejin Lee Koean J. Math. 23 (2015), No. 3, pp. 427 438 http://dx.doi.og/10.11568/kjm.2015.23.3.427 THE JEU DE TAQUIN ON THE SHIFTED RIM HOOK TABLEAUX Jaejin Lee Abstact. The Schensted algoithm fist descibed by Robinson

More information

Chapter 1: Introduction to Polar Coordinates

Chapter 1: Introduction to Polar Coordinates Habeman MTH Section III: ola Coodinates and Comple Numbes Chapte : Intoduction to ola Coodinates We ae all comfotable using ectangula (i.e., Catesian coodinates to descibe points on the plane. Fo eample,

More information

Solution to HW 3, Ma 1a Fall 2016

Solution to HW 3, Ma 1a Fall 2016 Solution to HW 3, Ma a Fall 206 Section 2. Execise 2: Let C be a subset of the eal numbes consisting of those eal numbes x having the popety that evey digit in the decimal expansion of x is, 3, 5, o 7.

More information

MATH 415, WEEK 3: Parameter-Dependence and Bifurcations

MATH 415, WEEK 3: Parameter-Dependence and Bifurcations MATH 415, WEEK 3: Paamete-Dependence and Bifucations 1 A Note on Paamete Dependence We should pause to make a bief note about the ole played in the study of dynamical systems by the system s paametes.

More information

The Erdős-Hajnal conjecture for rainbow triangles

The Erdős-Hajnal conjecture for rainbow triangles The Edős-Hajnal conjectue fo ainbow tiangles Jacob Fox Andey Ginshpun János Pach Abstact We pove that evey 3-coloing of the edges of the complete gaph on n vetices without a ainbow tiangle contains a set

More information

LECTURE 12. Special Solutions of Laplace's Equation. 1. Separation of Variables with Respect to Cartesian Coordinates. Suppose.

LECTURE 12. Special Solutions of Laplace's Equation. 1. Separation of Variables with Respect to Cartesian Coordinates. Suppose. 50 LECTURE 12 Special Solutions of Laplace's Equation 1. Sepaation of Vaiables with Respect to Catesian Coodinates Suppose è12.1è èx; yè =XèxèY èyè is a solution of è12.2è Then we must have è12.3è 2 x

More information

RADIAL POSITIVE SOLUTIONS FOR A NONPOSITONE PROBLEM IN AN ANNULUS

RADIAL POSITIVE SOLUTIONS FOR A NONPOSITONE PROBLEM IN AN ANNULUS Electonic Jounal of Diffeential Equations, Vol. 04 (04), o. 9, pp. 0. ISS: 07-669. UL: http://ejde.math.txstate.edu o http://ejde.math.unt.edu ftp ejde.math.txstate.edu ADIAL POSITIVE SOLUTIOS FO A OPOSITOE

More information

15.081J/6.251J Introduction to Mathematical Programming. Lecture 6: The Simplex Method II

15.081J/6.251J Introduction to Mathematical Programming. Lecture 6: The Simplex Method II 15081J/6251J Intoduction to Mathematical Pogamming ectue 6: The Simplex Method II 1 Outline Revised Simplex method Slide 1 The full tableau implementation Anticycling 2 Revised Simplex Initial data: A,

More information

Physics 107 TUTORIAL ASSIGNMENT #8

Physics 107 TUTORIAL ASSIGNMENT #8 Physics 07 TUTORIAL ASSIGNMENT #8 Cutnell & Johnson, 7 th edition Chapte 8: Poblems 5,, 3, 39, 76 Chapte 9: Poblems 9, 0, 4, 5, 6 Chapte 8 5 Inteactive Solution 8.5 povides a model fo solving this type

More information

3.6 Applied Optimization

3.6 Applied Optimization .6 Applied Optimization Section.6 Notes Page In this section we will be looking at wod poblems whee it asks us to maimize o minimize something. Fo all the poblems in this section you will be taking the

More information

Related Rates - the Basics

Related Rates - the Basics Related Rates - the Basics In this section we exploe the way we can use deivatives to find the velocity at which things ae changing ove time. Up to now we have been finding the deivative to compae the

More information

A Crash Course in (2 2) Matrices

A Crash Course in (2 2) Matrices A Cash Couse in ( ) Matices Seveal weeks woth of matix algeba in an hou (Relax, we will only stuy the simplest case, that of matices) Review topics: What is a matix (pl matices)? A matix is a ectangula

More information

Introduction Common Divisors. Discrete Mathematics Andrei Bulatov

Introduction Common Divisors. Discrete Mathematics Andrei Bulatov Intoduction Common Divisos Discete Mathematics Andei Bulatov Discete Mathematics Common Divisos 3- Pevious Lectue Integes Division, popeties of divisibility The division algoithm Repesentation of numbes

More information

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum 2. Electostatics D. Rakhesh Singh Kshetimayum 1 2.1 Intoduction In this chapte, we will study how to find the electostatic fields fo vaious cases? fo symmetic known chage distibution fo un-symmetic known

More information

Galois points on quartic surfaces

Galois points on quartic surfaces J. Math. Soc. Japan Vol. 53, No. 3, 2001 Galois points on quatic sufaces By Hisao Yoshihaa (Received Nov. 29, 1999) (Revised Ma. 30, 2000) Abstact. Let S be a smooth hypesuface in the pojective thee space

More information

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3.

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3. Appendix A Vecto Algeba As is natual, ou Aeospace Stuctues will be descibed in a Euclidean thee-dimensional space R 3. A.1 Vectos A vecto is used to epesent quantities that have both magnitude and diection.

More information

When two numbers are written as the product of their prime factors, they are in factored form.

When two numbers are written as the product of their prime factors, they are in factored form. 10 1 Study Guide Pages 420 425 Factos Because 3 4 12, we say that 3 and 4 ae factos of 12. In othe wods, factos ae the numbes you multiply to get a poduct. Since 2 6 12, 2 and 6 ae also factos of 12. The

More information

Target Boards, JEE Main & Advanced (IIT), NEET Physics Gauss Law. H. O. D. Physics, Concept Bokaro Centre P. K. Bharti

Target Boards, JEE Main & Advanced (IIT), NEET Physics Gauss Law. H. O. D. Physics, Concept Bokaro Centre P. K. Bharti Page 1 CONCPT: JB-, Nea Jitenda Cinema, City Cente, Bokao www.vidyadishti.og Gauss Law Autho: Panjal K. Bhati (IIT Khaagpu) Mb: 74884484 Taget Boads, J Main & Advanced (IIT), NT 15 Physics Gauss Law Autho:

More information

Connectedness of Ordered Rings of Fractions of C(X) with the m-topology

Connectedness of Ordered Rings of Fractions of C(X) with the m-topology Filomat 31:18 (2017), 5685 5693 https://doi.og/10.2298/fil1718685s Published by Faculty o Sciences and Mathematics, Univesity o Niš, Sebia Available at: http://www.pm.ni.ac.s/ilomat Connectedness o Odeed

More information

B. Spherical Wave Propagation

B. Spherical Wave Propagation 11/8/007 Spheical Wave Popagation notes 1/1 B. Spheical Wave Popagation Evey antenna launches a spheical wave, thus its powe density educes as a function of 1, whee is the distance fom the antenna. We

More information

PHYS 705: Classical Mechanics. Small Oscillations

PHYS 705: Classical Mechanics. Small Oscillations PHYS 705: Classical Mechanics Small Oscillations Fomulation of the Poblem Assumptions: V q - A consevative system with depending on position only - The tansfomation equation defining does not dep on time

More information

Output-Sensitive Algorithms for Computing Nearest-Neighbour Decision Boundaries

Output-Sensitive Algorithms for Computing Nearest-Neighbour Decision Boundaries Output-Sensitive Algoithms fo Computing Neaest-Neighbou Decision Boundaies David Bemne 1, Eik Demaine 2, Jeff Eickson 3, John Iacono 4, Stefan Langeman 5, Pat Moin 6, and Godfied Toussaint 7 1 Faculty

More information

Relating Branching Program Size and. Formula Size over the Full Binary Basis. FB Informatik, LS II, Univ. Dortmund, Dortmund, Germany

Relating Branching Program Size and. Formula Size over the Full Binary Basis. FB Informatik, LS II, Univ. Dortmund, Dortmund, Germany Relating Banching Pogam Size and omula Size ove the ull Binay Basis Matin Saueho y Ingo Wegene y Ralph Wechne z y B Infomatik, LS II, Univ. Dotmund, 44 Dotmund, Gemany z ankfut, Gemany sauehof/wegene@ls.cs.uni-dotmund.de

More information

Magnetic Field. Conference 6. Physics 102 General Physics II

Magnetic Field. Conference 6. Physics 102 General Physics II Physics 102 Confeence 6 Magnetic Field Confeence 6 Physics 102 Geneal Physics II Monday, Mach 3d, 2014 6.1 Quiz Poblem 6.1 Think about the magnetic field associated with an infinite, cuent caying wie.

More information

3.2 Centripetal Acceleration

3.2 Centripetal Acceleration unifom cicula motion the motion of an object with onstant speed along a cicula path of constant adius 3.2 Centipetal Acceleation The hamme thow is a tack-and-field event in which an athlete thows a hamme

More information

Additive Approximation for Edge-Deletion Problems

Additive Approximation for Edge-Deletion Problems Additive Appoximation fo Edge-Deletion Poblems Noga Alon Asaf Shapia Benny Sudakov Abstact A gaph popety is monotone if it is closed unde emoval of vetices and edges. In this pape we conside the following

More information

Math Section 4.2 Radians, Arc Length, and Area of a Sector

Math Section 4.2 Radians, Arc Length, and Area of a Sector Math 1330 - Section 4. Radians, Ac Length, and Aea of a Secto The wod tigonomety comes fom two Geek oots, tigonon, meaning having thee sides, and mete, meaning measue. We have aleady defined the six basic

More information

Energy Levels Of Hydrogen Atom Using Ladder Operators. Ava Khamseh Supervisor: Dr. Brian Pendleton The University of Edinburgh August 2011

Energy Levels Of Hydrogen Atom Using Ladder Operators. Ava Khamseh Supervisor: Dr. Brian Pendleton The University of Edinburgh August 2011 Enegy Levels Of Hydogen Atom Using Ladde Opeatos Ava Khamseh Supeviso: D. Bian Pendleton The Univesity of Edinbugh August 11 1 Abstact The aim of this pape is to fist use the Schödinge wavefunction methods

More information

(A) 2log( tan cot ) [ ], 2 MATHEMATICS. 1. Which of the following is correct?

(A) 2log( tan cot ) [ ], 2 MATHEMATICS. 1. Which of the following is correct? MATHEMATICS. Which of the following is coect? A L.P.P always has unique solution Evey L.P.P has an optimal solution A L.P.P admits two optimal solutions If a L.P.P admits two optimal solutions then it

More information

Motion in One Dimension

Motion in One Dimension Motion in One Dimension Intoduction: In this lab, you will investigate the motion of a olling cat as it tavels in a staight line. Although this setup may seem ovesimplified, you will soon see that a detailed

More information

We give improved upper bounds for the number of primitive solutions of the Thue inequality

We give improved upper bounds for the number of primitive solutions of the Thue inequality NUMBER OF SOLUTIONS OF CUBIC THUE INEQUALITIES WITH POSITIVE DISCRIMINANT N SARADHA AND DIVYUM SHARMA Abstact Let F(X, Y) be an ieducible binay cubic fom with intege coefficients and positive disciminant

More information

Reduced Implicant Tries

Reduced Implicant Tries Reduced Implicant Ties Technical Repot SUNYA-CS-07-01 Novembe, 2007 Neil V. Muay Depatment of Compute Science Univesity at Albany Albany, NY 12222 email: nvm@cs.albany.edu Eik Rosenthal Depatment of Mathematics

More information

Radian and Degree Measure

Radian and Degree Measure CHAT Pe-Calculus Radian and Degee Measue *Tigonomety comes fom the Geek wod meaning measuement of tiangles. It pimaily dealt with angles and tiangles as it petained to navigation, astonomy, and suveying.

More information

Lecture 7: Angular Momentum, Hydrogen Atom

Lecture 7: Angular Momentum, Hydrogen Atom Lectue 7: Angula Momentum, Hydogen Atom Vecto Quantization of Angula Momentum and Nomalization of 3D Rigid Roto wavefunctions Conside l, so L 2 2 2. Thus, we have L 2. Thee ae thee possibilities fo L z

More information

COMP Parallel Computing SMM (3) OpenMP Case Study: The Barnes-Hut N-body Algorithm

COMP Parallel Computing SMM (3) OpenMP Case Study: The Barnes-Hut N-body Algorithm COMP 633 - Paallel Computing Lectue 8 Septembe 14, 2017 SMM (3) OpenMP Case Study: The Banes-Hut N-body Algoithm Topics Case study: the Banes-Hut algoithm Study an impotant algoithm in scientific computing»

More information

Generalisations of a Four-Square Theorem. Hiroshi Okumura and John F. Rigby. results on a wooden board and dedicated it to a shrine or a temple.

Generalisations of a Four-Square Theorem. Hiroshi Okumura and John F. Rigby. results on a wooden board and dedicated it to a shrine or a temple. 20 enealisations of a ousquae Theoem Hioshi Okumua and John. Rigby In the 17 th {19 th centuies, Japanese people often wote thei mathematical esults on a wooden boad and dedicated it to a shine o a temple.

More information

MULTILAYER PERCEPTRONS

MULTILAYER PERCEPTRONS Last updated: Nov 26, 2012 MULTILAYER PERCEPTRONS Outline 2 Combining Linea Classifies Leaning Paametes Outline 3 Combining Linea Classifies Leaning Paametes Implementing Logical Relations 4 AND and OR

More information

Numerical Integration

Numerical Integration MCEN 473/573 Chapte 0 Numeical Integation Fall, 2006 Textbook, 0.4 and 0.5 Isopaametic Fomula Numeical Integation [] e [ ] T k = h B [ D][ B] e B Jdsdt In pactice, the element stiffness is calculated numeically.

More information

Non-Linear Dynamics Homework Solutions Week 2

Non-Linear Dynamics Homework Solutions Week 2 Non-Linea Dynamics Homewok Solutions Week Chis Small Mach, 7 Please email me at smach9@evegeen.edu with any questions o concens eguading these solutions. Fo the ececises fom section., we sketch all qualitatively

More information

Physics 111 Lecture 5 Circular Motion

Physics 111 Lecture 5 Circular Motion Physics 111 Lectue 5 Cicula Motion D. Ali ÖVGÜN EMU Physics Depatment www.aovgun.com Multiple Objects q A block of mass m1 on a ough, hoizontal suface is connected to a ball of mass m by a lightweight

More information

NOTE. Some New Bounds for Cover-Free Families

NOTE. Some New Bounds for Cover-Free Families Jounal of Combinatoial Theoy, Seies A 90, 224234 (2000) doi:10.1006jcta.1999.3036, available online at http:.idealibay.com on NOTE Some Ne Bounds fo Cove-Fee Families D. R. Stinson 1 and R. Wei Depatment

More information

Boundary Layers and Singular Perturbation Lectures 16 and 17 Boundary Layers and Singular Perturbation. x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9.

Boundary Layers and Singular Perturbation Lectures 16 and 17 Boundary Layers and Singular Perturbation. x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9. Lectues 16 and 17 Bounday Layes and Singula Petubation A Regula Petubation In some physical poblems, the solution is dependent on a paamete K. When the paamete K is vey small, it is natual to expect that

More information

Vectors, Vector Calculus, and Coordinate Systems

Vectors, Vector Calculus, and Coordinate Systems ! Revised Apil 11, 2017 1:48 PM! 1 Vectos, Vecto Calculus, and Coodinate Systems David Randall Physical laws and coodinate systems Fo the pesent discussion, we define a coodinate system as a tool fo descibing

More information

ASTR415: Problem Set #6

ASTR415: Problem Set #6 ASTR45: Poblem Set #6 Cuan D. Muhlbege Univesity of Mayland (Dated: May 7, 27) Using existing implementations of the leapfog and Runge-Kutta methods fo solving coupled odinay diffeential equations, seveal

More information

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam)

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam) (Sample 3) Exam 1 - Physics 202 - Patel SPRING 1998 FORM CODE - A (solution key at end of exam) Be sue to fill in you student numbe and FORM lette (A, B, C) on you answe sheet. If you foget to include

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II September 15, 2012 Prof. Alan Guth PROBLEM SET 2

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II September 15, 2012 Prof. Alan Guth PROBLEM SET 2 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Depatment Physics 8.07: Electomagnetism II Septembe 5, 202 Pof. Alan Guth PROBLEM SET 2 DUE DATE: Monday, Septembe 24, 202. Eithe hand it in at the lectue,

More information

Efficiency Loss in a Network Resource Allocation Game

Efficiency Loss in a Network Resource Allocation Game Efficiency Loss in a Netwok Resouce Allocation Game Ramesh Johai johai@mit.edu) John N. Tsitsiklis jnt@mit.edu) June 11, 2004 Abstact We exploe the popeties of a congestion game whee uses of a congested

More information

Seidel s Trapezoidal Partitioning Algorithm

Seidel s Trapezoidal Partitioning Algorithm CS68: Geometic Agoithms Handout #6 Design and Anaysis Oigina Handout #6 Stanfod Univesity Tuesday, 5 Febuay 99 Oigina Lectue #7: 30 Januay 99 Topics: Seide s Tapezoida Patitioning Agoithm Scibe: Michae

More information

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism Physics 2020, Sping 2005 Lab 5 page 1 of 8 Lab 5. Magnetism PART I: INTRODUCTION TO MAGNETS This week we will begin wok with magnets and the foces that they poduce. By now you ae an expet on setting up

More information

ON SPARSELY SCHEMMEL TOTIENT NUMBERS. Colin Defant 1 Department of Mathematics, University of Florida, Gainesville, Florida

ON SPARSELY SCHEMMEL TOTIENT NUMBERS. Colin Defant 1 Department of Mathematics, University of Florida, Gainesville, Florida #A8 INTEGERS 5 (205) ON SPARSEL SCHEMMEL TOTIENT NUMBERS Colin Defant Depatment of Mathematics, Univesity of Floida, Gainesville, Floida cdefant@ufl.edu Received: 7/30/4, Revised: 2/23/4, Accepted: 4/26/5,

More information

0606 ADDITIONAL MATHEMATICS

0606 ADDITIONAL MATHEMATICS UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Intenational Geneal Cetificate of Seconday Education MARK SCHEME fo the Octobe/Novembe 011 question pape fo the guidance of teaches 0606 ADDITIONAL MATHEMATICS

More information

COMPUTATIONS OF ELECTROMAGNETIC FIELDS RADIATED FROM COMPLEX LIGHTNING CHANNELS

COMPUTATIONS OF ELECTROMAGNETIC FIELDS RADIATED FROM COMPLEX LIGHTNING CHANNELS Pogess In Electomagnetics Reseach, PIER 73, 93 105, 2007 COMPUTATIONS OF ELECTROMAGNETIC FIELDS RADIATED FROM COMPLEX LIGHTNING CHANNELS T.-X. Song, Y.-H. Liu, and J.-M. Xiong School of Mechanical Engineeing

More information

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4! or. r ˆ = points from source q to observer Physics 8.0 Quiz One Equations Fall 006 F = 1 4" o q 1 q = q q ˆ 3 4" o = E 4" o ˆ = points fom souce q to obseve 1 dq E = # ˆ 4" 0 V "## E "d A = Q inside closed suface o d A points fom inside to V =

More information

ME 425: Aerodynamics

ME 425: Aerodynamics ME 5: Aeodynamics D ABM Toufique Hasan Pofesso Depatment of Mechanical Engineeing, BUET Lectue- 8 Apil 7 teachebuetacbd/toufiquehasan/ toufiquehasan@mebuetacbd ME5: Aeodynamics (Jan 7) Flow ove a stationay

More information

( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.

( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II. Homewok #8. hapte 7 Magnetic ields. 6 Eplain how ou would modif Gauss s law if scientists discoveed that single, isolated magnetic poles actuall eisted. Detemine the oncept Gauss law fo magnetism now eads

More information

4. Electrodynamic fields

4. Electrodynamic fields 4. Electodynamic fields D. Rakhesh Singh Kshetimayum 1 4.1 Intoduction Electodynamics Faaday s law Maxwell s equations Wave equations Lenz s law Integal fom Diffeential fom Phaso fom Bounday conditions

More information

SPECTRAL SEQUENCES: FRIEND OR FOE?

SPECTRAL SEQUENCES: FRIEND OR FOE? SPECTRAL SEQUENCES: FRIEND OR FOE? RAVI VAKIL Spectal sequences ae a poweful book-keeping tool fo poving things involving complicated commutative diagams. They wee intoduced by Leay in the 94 s at the

More information

Gauss Law. Physics 231 Lecture 2-1

Gauss Law. Physics 231 Lecture 2-1 Gauss Law Physics 31 Lectue -1 lectic Field Lines The numbe of field lines, also known as lines of foce, ae elated to stength of the electic field Moe appopiately it is the numbe of field lines cossing

More information

This brief note explains why the Michel-Levy colour chart for birefringence looks like this...

This brief note explains why the Michel-Levy colour chart for birefringence looks like this... This bief note explains why the Michel-Levy colou chat fo biefingence looks like this... Theoy of Levy Colou Chat fo Biefingent Mateials Between Cossed Polas Biefingence = n n, the diffeence of the efactive

More information

An Exact Solution of Navier Stokes Equation

An Exact Solution of Navier Stokes Equation An Exact Solution of Navie Stokes Equation A. Salih Depatment of Aeospace Engineeing Indian Institute of Space Science and Technology, Thiuvananthapuam, Keala, India. July 20 The pincipal difficulty in

More information

AP Physics C: Electricity and Magnetism 2001 Scoring Guidelines

AP Physics C: Electricity and Magnetism 2001 Scoring Guidelines AP Physics C: Electicity and Magnetism 1 Scoing Guidelines The mateials included in these files ae intended fo non-commecial use by AP teaches fo couse and exam pepaation; pemission fo any othe use must

More information

GROWTH ESTIMATES THROUGH SCALING FOR QUASILINEAR PARTIAL DIFFERENTIAL EQUATIONS

GROWTH ESTIMATES THROUGH SCALING FOR QUASILINEAR PARTIAL DIFFERENTIAL EQUATIONS Annales Academiæ Scientiaum Fennicæ Mathematica Volumen 32, 2007, 595 599 GROWTH ESTIMATES THROUGH SCALING FOR QUASILINEAR PARTIAL DIFFERENTIAL EQUATIONS Teo Kilpeläinen, Henik Shahgholian and Xiao Zhong

More information

New lower bounds for the independence number of sparse graphs and hypergraphs

New lower bounds for the independence number of sparse graphs and hypergraphs New lowe bounds fo the independence numbe of spase gaphs and hypegaphs Kunal Dutta, Dhuv Mubayi, and C.R. Subamanian May 23, 202 Abstact We obtain new lowe bounds fo the independence numbe of K -fee gaphs

More information

Section 8.2 Polar Coordinates

Section 8.2 Polar Coordinates Section 8. Pola Coodinates 467 Section 8. Pola Coodinates The coodinate system we ae most familia with is called the Catesian coodinate system, a ectangula plane divided into fou quadants by the hoizontal

More information

rt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t)

rt () is constant. We know how to find the length of the radius vector by r( t) r( t) r( t) Cicula Motion Fom ancient times cicula tajectoies hae occupied a special place in ou model of the Uniese. Although these obits hae been eplaced by the moe geneal elliptical geomety, cicula motion is still

More information

Goodness-of-fit for composite hypotheses.

Goodness-of-fit for composite hypotheses. Section 11 Goodness-of-fit fo composite hypotheses. Example. Let us conside a Matlab example. Let us geneate 50 obsevations fom N(1, 2): X=nomnd(1,2,50,1); Then, unning a chi-squaed goodness-of-fit test

More information

Functions Defined on Fuzzy Real Numbers According to Zadeh s Extension

Functions Defined on Fuzzy Real Numbers According to Zadeh s Extension Intenational Mathematical Foum, 3, 2008, no. 16, 763-776 Functions Defined on Fuzzy Real Numbes Accoding to Zadeh s Extension Oma A. AbuAaqob, Nabil T. Shawagfeh and Oma A. AbuGhneim 1 Mathematics Depatment,

More information

Dynamic Visualization of Complex Integrals with Cabri II Plus

Dynamic Visualization of Complex Integrals with Cabri II Plus Dynamic Visualiation of omplex Integals with abi II Plus Sae MIKI Kawai-juu, IES Japan Email: sand_pictue@hotmailcom Abstact: Dynamic visualiation helps us undestand the concepts of mathematics This pape

More information

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006 1 Qualifying Examination Electicity and Magnetism Solutions Januay 12, 2006 PROBLEM EA. a. Fist, we conside a unit length of cylinde to find the elationship between the total chage pe unit length λ and

More information

1 Notes on Order Statistics

1 Notes on Order Statistics 1 Notes on Ode Statistics Fo X a andom vecto in R n with distibution F, and π S n, define X π by and F π by X π (X π(1),..., X π(n) ) F π (x 1,..., x n ) F (x π 1 (1),..., x π 1 (n)); then the distibution

More information

Class #16 Monday, March 20, 2017

Class #16 Monday, March 20, 2017 D. Pogo Class #16 Monday, Mach 0, 017 D Non-Catesian Coodinate Systems A point in space can be specified by thee numbes:, y, and z. O, it can be specified by 3 diffeent numbes:,, and z, whee = cos, y =

More information

n 1 Cov(X,Y)= ( X i- X )( Y i-y ). N-1 i=1 * If variable X and variable Y tend to increase together, then c(x,y) > 0

n 1 Cov(X,Y)= ( X i- X )( Y i-y ). N-1 i=1 * If variable X and variable Y tend to increase together, then c(x,y) > 0 Covaiance and Peason Coelation Vatanian, SW 540 Both covaiance and coelation indicate the elationship between two (o moe) vaiables. Neithe the covaiance o coelation give the slope between the X and Y vaiable,

More information

Topic 4a Introduction to Root Finding & Bracketing Methods

Topic 4a Introduction to Root Finding & Bracketing Methods /8/18 Couse Instucto D. Raymond C. Rumpf Office: A 337 Phone: (915) 747 6958 E Mail: cumpf@utep.edu Topic 4a Intoduction to Root Finding & Backeting Methods EE 4386/531 Computational Methods in EE Outline

More information

Conspiracy and Information Flow in the Take-Grant Protection Model

Conspiracy and Information Flow in the Take-Grant Protection Model Conspiacy and Infomation Flow in the Take-Gant Potection Model Matt Bishop Depatment of Compute Science Univesity of Califonia at Davis Davis, CA 95616-8562 ABSTRACT The Take Gant Potection Model is a

More information

The 33 rd Balkan Mathematical Olympiad Tirana, May 7, 2016

The 33 rd Balkan Mathematical Olympiad Tirana, May 7, 2016 Language: English The 33 rd Balkan Mathematical Olympiad Tirana, May 7, 2016 Problem 1. Find all injective functions f : R R such that for every real number x and every positive integer n, n i f(x + i

More information

Radian Measure CHAPTER 5 MODELLING PERIODIC FUNCTIONS

Radian Measure CHAPTER 5 MODELLING PERIODIC FUNCTIONS 5.4 Radian Measue So fa, ou hae measued angles in degees, with 60 being one eolution aound a cicle. Thee is anothe wa to measue angles called adian measue. With adian measue, the ac length of a cicle is

More information

CALCULUS II Vectors. Paul Dawkins

CALCULUS II Vectors. Paul Dawkins CALCULUS II Vectos Paul Dawkins Table of Contents Peface... ii Vectos... 3 Intoduction... 3 Vectos The Basics... 4 Vecto Aithmetic... 8 Dot Poduct... 13 Coss Poduct... 21 2007 Paul Dawkins i http://tutoial.math.lama.edu/tems.aspx

More information

Physics 111 Lecture 5 (Walker: 3.3-6) Vectors & Vector Math Motion Vectors Sept. 11, 2009

Physics 111 Lecture 5 (Walker: 3.3-6) Vectors & Vector Math Motion Vectors Sept. 11, 2009 Physics 111 Lectue 5 (Walke: 3.3-6) Vectos & Vecto Math Motion Vectos Sept. 11, 2009 Quiz Monday - Chap. 2 1 Resolving a vecto into x-component & y- component: Pola Coodinates Catesian Coodinates x y =

More information

A Hartree-Fock Example Using Helium

A Hartree-Fock Example Using Helium Univesity of Connecticut DigitalCommons@UConn Chemisty Education Mateials Depatment of Chemisty June 6 A Hatee-Fock Example Using Helium Cal W. David Univesity of Connecticut, Cal.David@uconn.edu Follow

More information

SPREADING SPEED AND PROFILE FOR NONLINEAR STEFAN PROBLEMS IN HIGH SPACE DIMENSIONS

SPREADING SPEED AND PROFILE FOR NONLINEAR STEFAN PROBLEMS IN HIGH SPACE DIMENSIONS SPREADING SPEED AND PROFILE FOR NONLINEAR STEFAN PROBLEMS IN HIGH SPACE DIMENSIONS YIHONG DU, HIROSHI MATSUZAWA AND MAOLIN ZHOU Abstact. We conside nonlinea diffusion poblems of the fom u t = u + f(u)

More information

Math Notes on Kepler s first law 1. r(t) kp(t)

Math Notes on Kepler s first law 1. r(t) kp(t) Math 7 - Notes on Keple s fist law Planetay motion and Keple s Laws We conside the motion of a single planet about the sun; fo simplicity, we assign coodinates in R 3 so that the position of the sun is

More information